04.2 New certificate changes evaluation

Content and key concepts

On the previous page: chance of multiple outcomes.
The positive signal is not equal to the certainty of the event: its original rareness is important; the Bayes formula links the new certificate to the previous evaluation.
Of 10,000 incidents, 100 have the right property.
Для изменения рисунка включитеJavaScript; расчёт и теория доступны далее.
Do not confuse the direction of the condition
For events A,B at P(B)>0, the determination of the imputation probability: P(A~B)=P(A\B)/P(B). Since the joint probability can be recorded by P(b\A)P(E) obtained by Bayes:
P
(
A
∣
B
)
=
P
(
B
∣
A
)
P
(
A
)
P
(
B
∣
A
)
P
(
A
)
+
P
(
B
∣
¬
A
)
P
(
¬
A
)
.
P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\neg A)P(\neg A)}.
P
(
A
∣
B
)
=
P
(
B
∣
A
)
P
(
A
)
+
P
(
B
∣
¬
A
)
P
(
¬
A
)
P
(
B
∣
A
)
P
(
A
)
​
.
It's an algebra over one event, not a particular mystical way of changing beliefs. M04 §3.5, §3.11. In lab A, it means a rare defect, B is a positive signal from a training detector. This is not a medical advice.
Let the prior=P(A)=01, sensitivity=P (B\A) = 9, false-positive rate=P[B\BA] = .05. Ten thousand sites are expected to have 100 defects, 90 true signals, and 495 false. Among the 585 signals, only 90 is defective: posterior=90/585=2/13 E.153846, not 9.
Experiment.
Leave the characteristics of the detector and increase the prior to the.5. Posterior will become.9/(9+.05)E.947368. The same detector has changed the composition of the flow. Therefore, "90 per cent sensitivity" does not determine the proportion of true positive results without the base frequency.
Easy to be wrong.
The numbers per 10,000 are mathematical expectations that are not guaranteed by the number of objects in each experience. P(A10/B) is not P(B)A. If P(b)=0, the imputation is not defined; the interface explains the absence of possible positive signals, does not show 0/0. The conditional connection does not prove the causal effect of the intervention.
prior, sensitivity, fpr = .01,.9,.05
positive = prior*sensitivity+(1-prior)*fpr
posterior = prior*sensitivity/positive
assert abs(posterior-2/13)<1e-12
The code repeats the operations just taken care of.
Completely checked Python-example Chapter
Available with a separate file; the short piece shown does not replace its implementation.
The frequent positive signal does not eliminate the rareness of the initial event. Direction of the condition is important.
What settings explain observations better. How does this question continue with the result?
Python examples of Chapter 04
Academic base